Question 59
Let $a$, $b$, $c$ be positive real numbers such that $abc = 1$. Show that
$$
\frac{1}{a^{3}+b c}+\frac{1}{b^{3}+c a}+\frac{1}{c^{3}+a b} \leq \frac{(a b+b c+c a)^{2}}{6}
$$
$$
\frac{1}{a^{3}+b c}+\frac{1}{b^{3}+c a}+\frac{1}{c^{3}+a b} \leq \frac{(a b+b c+c a)^{2}}{6}
$$